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To find the equation of the normal to the curve [tex]\( y = 5x^3 - 6x^2 + 2x + 5 \)[/tex] at the point [tex]\((1, 2)\)[/tex], we will go through the following steps:
1. Find the derivative of the function to get the slope of the tangent line at any point [tex]\( x \)[/tex]. The derivative of the function [tex]\( y = 5x^3 - 6x^2 + 2x + 5 \)[/tex] with respect to [tex]\( x \)[/tex] is:
[tex]\[ y' = \frac{d}{dx}(5x^3 - 6x^2 + 2x + 5) = 15x^2 - 12x + 2 \][/tex]
2. Evaluate the derivative at the given point [tex]\( x = 1 \)[/tex] to find the slope of the tangent line at [tex]\((1, 2)\)[/tex]:
[tex]\[ y'(1) = 15(1)^2 - 12(1) + 2 = 15 - 12 + 2 = 5 \][/tex]
Therefore, the slope of the tangent line at [tex]\((1, 2)\)[/tex] is [tex]\( 5 \)[/tex].
3. Find the slope of the normal line. The normal line is perpendicular to the tangent line. The slope of a line that is perpendicular to a line with slope [tex]\( m \)[/tex] is the negative reciprocal of [tex]\( m \)[/tex]. Hence, the slope of the normal line to the tangent with slope [tex]\( 5 \)[/tex] is:
[tex]\[ \text{slope of normal line} = -\frac{1}{5} \][/tex]
4. Use the point-slope form of the line equation to find the equation of the normal line. The point-slope form is given by:
[tex]\[ y - y_1 = m(x - x_1) \][/tex]
where [tex]\((x_1, y_1)\)[/tex] is the point [tex]\((1, 2)\)[/tex] and [tex]\( m \)[/tex] is the slope of the normal line [tex]\(-\frac{1}{5}\)[/tex]. Substituting these values in, we get:
[tex]\[ y - 2 = -\frac{1}{5}(x - 1) \][/tex]
Simplifying this equation:
[tex]\[ y - 2 = -\frac{1}{5}x + \frac{1}{5} \][/tex]
Adding 2 to both sides:
[tex]\[ y = -\frac{1}{5}x + \frac{1}{5} + 2 \][/tex]
Combining the constants on the right-hand side:
[tex]\[ y = -\frac{1}{5}x + \frac{1}{5} + \frac{10}{5} \][/tex]
[tex]\[ y = -\frac{1}{5}x + \frac{11}{5} \][/tex]
Therefore, the equation of the normal line to the curve [tex]\( y = 5x^3 - 6x^2 + 2x + 5 \)[/tex] at the point [tex]\((1, 2)\)[/tex] is:
[tex]\[ y = -\frac{1}{5}x + \frac{11}{5} \][/tex]
For further simplification, this can also be written as:
[tex]\[ \frac{11}{5} - \frac{1}{5}x = y \][/tex]
Or, equivalently:
[tex]\[ \boxed{\frac{11}{5} - \frac{x}{5} = y} \][/tex]
1. Find the derivative of the function to get the slope of the tangent line at any point [tex]\( x \)[/tex]. The derivative of the function [tex]\( y = 5x^3 - 6x^2 + 2x + 5 \)[/tex] with respect to [tex]\( x \)[/tex] is:
[tex]\[ y' = \frac{d}{dx}(5x^3 - 6x^2 + 2x + 5) = 15x^2 - 12x + 2 \][/tex]
2. Evaluate the derivative at the given point [tex]\( x = 1 \)[/tex] to find the slope of the tangent line at [tex]\((1, 2)\)[/tex]:
[tex]\[ y'(1) = 15(1)^2 - 12(1) + 2 = 15 - 12 + 2 = 5 \][/tex]
Therefore, the slope of the tangent line at [tex]\((1, 2)\)[/tex] is [tex]\( 5 \)[/tex].
3. Find the slope of the normal line. The normal line is perpendicular to the tangent line. The slope of a line that is perpendicular to a line with slope [tex]\( m \)[/tex] is the negative reciprocal of [tex]\( m \)[/tex]. Hence, the slope of the normal line to the tangent with slope [tex]\( 5 \)[/tex] is:
[tex]\[ \text{slope of normal line} = -\frac{1}{5} \][/tex]
4. Use the point-slope form of the line equation to find the equation of the normal line. The point-slope form is given by:
[tex]\[ y - y_1 = m(x - x_1) \][/tex]
where [tex]\((x_1, y_1)\)[/tex] is the point [tex]\((1, 2)\)[/tex] and [tex]\( m \)[/tex] is the slope of the normal line [tex]\(-\frac{1}{5}\)[/tex]. Substituting these values in, we get:
[tex]\[ y - 2 = -\frac{1}{5}(x - 1) \][/tex]
Simplifying this equation:
[tex]\[ y - 2 = -\frac{1}{5}x + \frac{1}{5} \][/tex]
Adding 2 to both sides:
[tex]\[ y = -\frac{1}{5}x + \frac{1}{5} + 2 \][/tex]
Combining the constants on the right-hand side:
[tex]\[ y = -\frac{1}{5}x + \frac{1}{5} + \frac{10}{5} \][/tex]
[tex]\[ y = -\frac{1}{5}x + \frac{11}{5} \][/tex]
Therefore, the equation of the normal line to the curve [tex]\( y = 5x^3 - 6x^2 + 2x + 5 \)[/tex] at the point [tex]\((1, 2)\)[/tex] is:
[tex]\[ y = -\frac{1}{5}x + \frac{11}{5} \][/tex]
For further simplification, this can also be written as:
[tex]\[ \frac{11}{5} - \frac{1}{5}x = y \][/tex]
Or, equivalently:
[tex]\[ \boxed{\frac{11}{5} - \frac{x}{5} = y} \][/tex]
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